11

KineticsKinetics

Chapter 15Chapter 15

22

The study of rxn ratesThe study of rxn rates

• Rxn rate = Rxn rate = concentration/concentration/timetime

• Example: Example: 2N2N22OO55 4NO 4NO22 + O + O22

• As reactant As reactant concentration concentration decreases decreases products’ products’ concentrations concentrations increaseincrease

2 5

2

2

- [N O ]rate of rxn =

t[NO ]

rate of rxn = t

[O ]rate of rxn =

t

33

Let’s do some calculationsLet’s do some calculations

• From 300s to 400s, From 300s to 400s,

--[N[N22OO55]/]/t = t =

-.0019M/100s-.0019M/100s

• What are What are [NO[NO22]/]/t & t & [O[O22]/]/t during the same t during the same time interval?time interval?

• Let’s work this out…Let’s work this out…

• Is rxn rate constant?Is rxn rate constant?

– Let’s look at Let’s look at [NO[NO22]/]/t t at different time at different time intervalsintervals

44

More…More…

• The previous rates The previous rates were were average ratesaverage rates

• Different from Different from instantaneous ratesinstantaneous rates (@ single pt)(@ single pt)

• Ex:Ex:[N[N22OO55]=-0.0019M ]=-0.0019M

((@ 200s)@ 200s)• Car speed is Car speed is

analogousanalogous

55

Continued Continued

• We can also We can also calculate changes calculate changes mathematically mathematically based on based on stoichiometric stoichiometric relationshipsrelationships

• Let’s seeLet’s see

• http://wps.prenhall.http://wps.prenhall.com/wps/media/objcom/wps/media/objects/167/172009/Dects/167/172009/DecompositionofN2OecompositionofN2O5.html5.html

66

Writing rate expressionWriting rate expression

2N2N22OO55 4NO 4NO22 + O + O22

• To equate rates of To equate rates of disappearance or disappearance or appearanceappearance– Divide the Divide the

stoichiometric stoichiometric coefficient in the coefficient in the balanced equationbalanced equation

2 5 2 2[N O ] [NO ] [O ]1 1 = + = +2 4t t t

77

Let’s work on thisLet’s work on this

• a) Give four related rate expressions for a) Give four related rate expressions for the rate of the following reaction:the rate of the following reaction:

2H2H22COCO(g)(g) + O + O2(g)2(g) 2CO 2CO(g)(g) + 2H + 2H22OO(g)(g)

• b) Give three related rate expressions b) Give three related rate expressions for the rate of the following reaction:for the rate of the following reaction:

NN2(g)2(g) + 3H + 3H2(g)2(g) 2NH 2NH3(g)3(g)

88

Let’s also work on this:Let’s also work on this:

• Consider the reaction:Consider the reaction: A + 2B A + 2B C C

• Give the three related Give the three related rate expressions for rate expressions for the rate of the the rate of the reaction.reaction.

• Using the following Using the following data, determine the data, determine the average rate of the average rate of the reaction, and the rate reaction, and the rate between 30 and 40 between 30 and 40 seconds.seconds.

• Time (s): 0.0, 10.0, Time (s): 0.0, 10.0, 20.0, 30.0, & 40.020.0, 30.0, & 40.0

• [A] (M): 1.000, 0.833, [A] (M): 1.000, 0.833, 0.714, 0.625, & 0.555, 0.714, 0.625, & 0.555, respectivelyrespectively

99

Reaction conditions and Reaction conditions and ratesrates• Higher temp Higher temp faster rxn rate faster rxn rate

– Raise in temp by 10Raise in temp by 10°C °C double reaction rate double reaction rate

• Higher concentration Higher concentration faster rxn rate faster rxn rate

• CatalystsCatalysts speed up rxn rates speed up rxn rates– They don’t reactThey don’t react– They lower the activation energyThey lower the activation energy

• Enzymes (proteins) in organismsEnzymes (proteins) in organisms

• Metals, salts, etc. in chemical rxnsMetals, salts, etc. in chemical rxns

• Question: What does acid rain do to enzymes?Question: What does acid rain do to enzymes?

1010

Concentration:Concentration:2N2N22OO55 4NO 4NO22 + O + O22

• If you double [NIf you double [N22OO55], you double the ], you double the rxn raterxn rate– Rate of rxn Rate of rxn [N [N22OO55]]

• Other rxns have different Other rxns have different relationships w/concentration-rxn relationships w/concentration-rxn ratesrates

1111

Rate equationsRate equations

• Describe relationship between reactant Describe relationship between reactant concentration and rxn rateconcentration and rxn rate

• For 2NFor 2N22OO55 4NO 4NO22 + O + O22

– Rate of rxn = k[NRate of rxn = k[N22OO55]]• k = rate constantk = rate constant

– Is the rate of rxn constant too?Is the rate of rxn constant too?• Generic expression for rate eq: aA + bB Generic expression for rate eq: aA + bB xX xX• So rate = k[A]So rate = k[A]mm[B][B]nn

– Where m & n need not equal stoichiometric ratios!Where m & n need not equal stoichiometric ratios!– They can be zero, fractions, even negative #’sThey can be zero, fractions, even negative #’s– Empirically verified valuesEmpirically verified values

1212

Order of rxn:Order of rxn:Exponent of its concentration Exponent of its concentration term for each itemterm for each item • 2NO2NO(g)(g) + Cl + Cl2(g)2(g) 2NOCl 2NOCl(g)(g) • Rxn rate for NO is Rxn rate for NO is second ordersecond order• Rxn rate for ClRxn rate for Cl22 is is first orderfirst order

– Thus, rxn rate = k[NO]Thus, rxn rate = k[NO]22[Cl[Cl22]]• Total order of rxn = Total order of rxn = summationsummation of of

exponents of all concentration itemsexponents of all concentration items– Therefore, overall rxn rate is Therefore, overall rxn rate is third orderthird order

• Remember, m & n need not equal Remember, m & n need not equal stoichiometric ratios!stoichiometric ratios!

1313

Reaction mechanismsReaction mechanisms

• Step by step chemical equations that, when Step by step chemical equations that, when summed, give net rxnsummed, give net rxn

• For ex: For ex:

BrBr2(g)2(g) + 2NO + 2NO(g)(g) 2BrNO 2BrNO(g)(g)

1) Br1) Br2(g) 2(g) + NO+ NO(g)(g) Br Br22NONO(g)(g)

2) Br2) Br22NONO(g) (g) + NO+ NO(g)(g) 2BrNO 2BrNO(g)(g)

• Each step:Each step:– Elementary stepElementary step

1414

MolecularityMolecularity

• Elementary steps Elementary steps classified according to # classified according to # of reactant molecules of reactant molecules – MolecularityMolecularity

• Unimolecular: Unimolecular: – A A products products – Rate expression: Rate expression: k[A]k[A]

• Bimolecular: Bimolecular: – A + B A + B products products– Rate expression: Rate expression: k[A]k[A]

[B][B]

• Termolecular:Termolecular:– A + B + C A + B + C products products

• (Or 3A or 2A + B, etc.)(Or 3A or 2A + B, etc.)

– Rate expression: Rate expression: k[A][B]k[A][B][C][C] (or (or k[A]k[A]33 or or k[A]k[A]22[B][B], , etc.)etc.)

• In elementary steps, In elementary steps, stoichiometry defines stoichiometry defines rate equation!rate equation!– K different for each stepK different for each step

1515

Rate-determining stepRate-determining step

• The slowest elementary step is the The slowest elementary step is the rate-limiting steprate-limiting step– Overall rxn follows this rate-limiting stepOverall rxn follows this rate-limiting step

•However, intermediate not included in final However, intermediate not included in final rate lawrate law

1616

An exampleAn example

2NO2NO(g)(g) + O + O2(g)2(g) → 2NO → 2NO2(g)2(g) • NONO(g)(g)

+ NO+ NO(g)(g) N N22OO2(g)2(g) (fast, eq.) (fast, eq.) • NN22OO2(g) 2(g) + O+ O2(g)2(g) → NO → NO2(g)2(g) (slow) (slow) • Using slow, rate-determining, step Using slow, rate-determining, step

– Rate = kRate = k22[N[N22OO22][O][O22]]• Since fast-step at eq. Since fast-step at eq.

– kk11[NO][NO]22 = k = k-1-1[N[N22OO22]]– Thus, K = kThus, K = k11/k/k-1 -1 = [N= [N22OO22]/[NO]]/[NO]22

• Intermediate ([NIntermediate ([N22OO22]) = K[NO]]) = K[NO]22

– Since rate = kSince rate = k22[N[N22OO22][O][O22]]•Rate = kRate = k2 2 K[NO]K[NO]22[O[O22]]

• Let’s say kLet’s say k2 2 K = k’K = k’ rate = k’[NO]rate = k’[NO]22[O[O22]]

1717

Another problemAnother problem

2H2H2(g)2(g) + 2NO + 2NO(g)(g) 2H 2H22OO(g)(g) + N + N2(g)2(g)

• 2NO2NO(g)(g) N N22OO2(g)2(g); fast; fast

• HH2(g)2(g) + N + N22OO2(g)2(g) H H22OO(g)(g) + N + N22OO(g)(g); slow; slow

• NN22OO(g)(g) + H + H2(g)2(g) N N2(g)2(g) + H + H22OO(g)(g); fast; fast

1818

Solution Solution (g) 2 2(g)

2(g) 2 2(g) 2 (g) 2 (g)

2 (g) 2(g) 2(g) 2 (g)

2fwd 1 rev -1 2 2

2fwd rev 1 -1 2 2

12 2

2NO N O ; fast

H + N O H O + N O ; slow

N O + H N + H O ; fast

fast: rate = k [NO] , rate = k [N O ]

Since rate rate , k [NO] k [N O ]

k [NO]Thus, [N O ]=

2

2

-1

22 2 2 2 2 2

22

K'[NO]k

slow: rate = k [H ][N O ]=k [H ] K'[NO]

rate = K"[H ][NO]

1919

Rate = changeRate = change

• 2NH2NH33 N N22 + 3H + 3H22

• Rate = k[NHRate = k[NH33]]00 = k = k– Zero order Zero order rxn rxn

raterate

• Any tinkering of Any tinkering of concentration will concentration will not change rxn not change rxn rate of speciesrate of species

• k = mol/(L k = mol/(L time) time)

2020

Rate = k[NO]Rate = k[NO]22[Cl[Cl22]]

• Rxn rate for ClRxn rate for Cl22 is is first orderfirst order

rate doubled rate doubled when [Clwhen [Cl22] doubled] doubled

• k = timek = time-1-1

2121

Rate = k[NO]Rate = k[NO]22[Cl[Cl22]]

• Rxn rate for NO is Rxn rate for NO is second ordersecond order

• Doubling [NO] Doubling [NO] quadruples (x4) rxn quadruples (x4) rxn raterate

• k = L/(mol k = L/(mol time) time)

2222

Let’s do this:Let’s do this:

• Nitrosyl bromide, NOBr, is formed from Nitrosyl bromide, NOBr, is formed from NO and BrNO and Br22..

2NO2NO(g) (g) + Br+ Br2(g)2(g) 2NOBr 2NOBr(g)(g)

• Experiment show that the reaction is first Experiment show that the reaction is first order in Brorder in Br22 and second order in NO. and second order in NO.

– Write the rate law for the reaction.Write the rate law for the reaction.• If the concentration of BrIf the concentration of Br22 is tripled, how is tripled, how

will the reaction rate change?will the reaction rate change?• What happens to the reaction rate when What happens to the reaction rate when

the concentration of NO is doubled?the concentration of NO is doubled?

2323

Determining rate equationsDetermining rate equations

• Done in the labDone in the lab

• Generally, calculated after 1-2% of Generally, calculated after 1-2% of limiting reactant consumedlimiting reactant consumed– Lessens chance of side-rxns throwing Lessens chance of side-rxns throwing

things offthings off

2424

Let’s work on this:Let’s work on this:

• The rate for the oxidation of iron (II) by cerium (IV) is The rate for the oxidation of iron (II) by cerium (IV) is measured at several different initial concentrations of the measured at several different initial concentrations of the two reactants:two reactants:

CeCe4+4+(aq)(aq) + Fe + Fe2+2+

(aq)(aq) Ce Ce3+3+(aq)(aq) + Fe + Fe3+3+

(aq)(aq)

• [Ce[Ce4+4+] (M): 1.1 x 10] (M): 1.1 x 10-5-5, 1.1 x 10, 1.1 x 10-5-5, 3.4 x 10, 3.4 x 10-5-5

• [Fe[Fe2+2+] (M): 1.8 x 10] (M): 1.8 x 10-5-5, 2.8 x 10, 2.8 x 10-5-5, 2.8 x 10, 2.8 x 10-5-5

• Rate (M/s): 2.0 x 10Rate (M/s): 2.0 x 10-7-7, 3.1 x 10, 3.1 x 10-7-7, 9.5 x 10, 9.5 x 10-7-7

– Write the rate law for this reaction, determining the Write the rate law for this reaction, determining the orders of the reaction with respect to Ce (IV) and Fe (II). orders of the reaction with respect to Ce (IV) and Fe (II). What is the overall order of the reaction? What is the overall order of the reaction?

– Calculate the rate constant, k, and give its units.Calculate the rate constant, k, and give its units.– Predict the reaction rate for a solution in which [CePredict the reaction rate for a solution in which [Ce4+4+] is ] is

2.6 x 102.6 x 10-5-5 M and [Fe M and [Fe2+2+] is 1.3 x 10] is 1.3 x 10-5-5 M. M.

2525

Concentration-time Concentration-time relationships: Integrated rate relationships: Integrated rate lawslaws• Zero-order rxnsZero-order rxns• -(-([R]/[R]/t) = k[R]t) = k[R]00

• Using integral Using integral calculus, calculus, integrated rate integrated rate equationequation::

k = concentration/timek = concentration/time

i f[R] - [R] = kt

2626

Concentration-time Concentration-time relationships: Integrated rate relationships: Integrated rate lawslaws• First-order rxnsFirst-order rxns• --[R]/[R]/t = k[R]t = k[R]

k = timek = time-1-1

f

i

[R]ln = -kt

[R]

2727

Integrated rate lawsIntegrated rate laws

• Second-order rxnsSecond-order rxns• --[R]/[R]/t = k[R]t = k[R]22

k=1/k=1/(concentration(concentrationtime)time)

f i

1 1 - = kt

[R] [R]

2828

Using graphs to solve rxn order Using graphs to solve rxn order and rate constantsand rate constants

• Zero-order: Zero-order:

• [R][R]ii – [R] – [R]ff = kt = kt

• Rearrange this in the form:Rearrange this in the form:y = mx + by = mx + b

2929

Using graphs to solve rxn order Using graphs to solve rxn order and rate constantsand rate constants• First-order:First-order:

• Rearrange this in the form:Rearrange this in the form:y = mx + by = mx + b

• Solve the following: 2H2O2(aq) 2H2O(l) + O2(g)

• The reaction is first order in peroxide, and the rate constant, k, for this reaction is 1.06 x 10-3/min.

• If the initial peroxide concentration is 0.020 M, what is the concentration after 135 min?

f

i

[R]ln = -kt

[R]

3030

Solution Solution

f

i

-3f

-3

f

4.06f

[R]ln = -kt

[R]

[R] -1.06 10ln = (135 mins)min[0.020 M]

-1.06 10ln[R] ln[0.020 M] (135 mins)min

[R] .017Me

3131

Using graphs to solve rxn order Using graphs to solve rxn order and rate constantsand rate constants• Second-order:Second-order:

• Rearrange this in the form:Rearrange this in the form:y = mx + by = mx + b

• The gas phase decomposition of HI into hydrogen and iodine is second order in HI.

• The rate constant for the reaction is 30/Mmin. • How long must one wait for the concentration of

HI to decrease from 0.010 M to 0.0050 M?

f i

1 1 - = kt

[R] [R]

3232

Solution Solution

f i

1 1 - = kt

[R] [R]

1 1 M - 30 tmin[0.0050M] [0.010M]

t=3.3 min

3333

Half-lifeHalf-life

• Time required for Time required for reactant reactant concentration to concentration to decrease to ½ initial decrease to ½ initial valuevalue– Longer half-life Longer half-life

means slower rxnmeans slower rxn– Usually used for Usually used for

11stst-order rxns (like -order rxns (like radioactive decay)radioactive decay)

f i

f

i

f1

2i

12

12

1[R] = [R]2[R]1

2 [R]

[R]1ln( ) ln( ) kt2 [R]

1ln( ) 0.693 kt2

0.693t =

k

3434

Half-lifeHalf-life

• For 1For 1stst-order rxns, -order rxns, tt1/21/2 is independent is independent of concof conc

3535

Half-life: 2Half-life: 2ndnd order rxns order rxns

• Half-life increases Half-life increases with decreasing with decreasing concentrationconcentration

• Derive this!Derive this!

12

i

1t =

k[R]

3636

Half-life: zero-order rxnsHalf-life: zero-order rxns

• Half-life decreases Half-life decreases with decreasing with decreasing concentrationconcentration

i1

2

[R]t

2k

3737

Problem Problem

• Sucrose decomposes to fructose and Sucrose decomposes to fructose and glucose in acid solution with the rate glucose in acid solution with the rate law of:law of:

Rate = k[sucrose]; k=0.208 hrRate = k[sucrose]; k=0.208 hr-1-1 @ @ 2525°C°C

• What amount of time is required for What amount of time is required for 87.5% of the initial concentration of 87.5% of the initial concentration of sucrose to decompose?sucrose to decompose?

3838

Solution Solution

[12.5M] -0.208ln = thr[100.0M]

t=1.00hr

3939

Collision TheoryCollision Theory

1. Reacting molecules must collide 1. Reacting molecules must collide with one another.with one another.

2. Reacting molecules must collide 2. Reacting molecules must collide with sufficient energy to break with sufficient energy to break bonds.bonds.

3. Molecules must collide in correct 3. Molecules must collide in correct orientation to form new species.orientation to form new species.

4040

Temperature Temperature

• Higher Higher temperatures lead temperatures lead to fast reactionsto fast reactions

• Molecules have Molecules have varying varying temperaturestemperatures– At higher At higher

temperatures, more temperatures, more molecules have molecules have higher energieshigher energies

4141

Activation energy and Activation energy and catalystscatalysts• Activation energy:Activation energy:

– Energy required to form Energy required to form productsproducts

• Methane + oxygen gasMethane + oxygen gas– Low activation energyLow activation energy

• Xenon + oxygen gasXenon + oxygen gas– Humongous activation Humongous activation

energyenergy

• Catalysts dramatically Catalysts dramatically reduce activation reduce activation energy w/out being energy w/out being consumedconsumed

4242

Arrhenius equationArrhenius equation• Includes tenets of collision Includes tenets of collision

theory; i.e., collision theory; i.e., collision frequency, temp (energy), frequency, temp (energy), and correct orientationand correct orientation

• k = Aek = Ae-E-Eaa/RT/RT

• A = frequency factor A = frequency factor (L/mol(L/mols)s)– Empirically derived Empirically derived

relationship between relationship between reaction rate and tempreaction rate and temp

• ee-E-Eaa/RT /RT = fraction of molecules = fraction of molecules w/min energy needed for rxn w/min energy needed for rxn (R = gas constant = 8.314 (R = gas constant = 8.314 J/molJ/molK)K)

• Can put equation in Can put equation in “y=mx+b” form“y=mx+b” form– How?How?

a-E 1ln k = + ln A

R T

4343

The Arrhenius equationThe Arrhenius equation

4444

Problem Problem

• Given:Given:y = -24371x + 28.204y = -24371x + 28.204

• Solve for ESolve for Eaa

4545

Solution Solution

a a

5a

y = -24371x + 28.204

E E-24371K = -

JR 8.314 mol KJE 2.0262 10 mol

4646

If given two points…If given two points…

a2

1 2 1

-Ek 1 1ln( ) = ( - )

k R T T